A238010 Number A(n,k) of partitions of k^n into parts that are at most n; square array A(n,k), n>=0, k>=0, read by antidiagonals.
0, 0, 1, 0, 1, 1, 0, 1, 1, 1, 0, 1, 3, 1, 1, 0, 1, 5, 10, 1, 1, 0, 1, 9, 75, 64, 1, 1, 0, 1, 13, 374, 4410, 831, 1, 1, 0, 1, 19, 1365, 123464, 1366617, 26207, 1, 1, 0, 1, 25, 3997, 1736385, 393073019, 2559274110, 2239706, 1, 1
Offset: 0
Examples
A(3,2) = 10: 332, 2222, 3221, 3311, 22211, 32111, 221111, 311111, 2111111, 11111111. A(2,3) = 5: 22221, 222111, 2211111, 21111111, 111111111. A(2,4) = 9: 22222222, 222222211, 2222221111, 22222111111, 222211111111, 2221111111111, 22111111111111, 211111111111111, 1111111111111111. Square array A(n,k) begins: 0, 0, 0, 0, 0, 0, ... 1, 1, 1, 1, 1, 1, ... 1, 1, 3, 5, 9, 13, ... 1, 1, 10, 75, 374, 1365, ... 1, 1, 64, 4410, 123464, 1736385, ... 1, 1, 831, 1366617, 393073019, 33432635477, ...
Links
- Alois P. Heinz, Antidiagonals n = 0..43, flattened
- A. V. Sills and D. Zeilberger, Formulae for the number of partitions of n into at most m parts (using the quasi-polynomial ansatz) (arXiv:1108.4391 [math.CO])
Crossrefs
Programs
-
Mathematica
A[n_, k_] := SeriesCoefficient[Product[1/(1-x^j), {j, 1, n}], {x, 0, k^n}]; A[0, 0] = 0; Table[A[n-k, k], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Feb 17 2017 *)
Formula
A(n,k) = [x^(k^n)] Product_{j=1..n} 1/(1-x^j).
Comments